Question

Find $\frac{dy}{dx}$ in each of the following cases:

$\sin xy+\cos \left(x+y\right)=1$

Answer 1

$\mathrm{We}\mathrm{have},\sin xy+\cos \left(x+y\right)=1$
Differentiating with respect to x, we get,
$$\begin{gathered} \frac{d}{dx}\left(\sin xy\right)+\frac{d}{dx}\cos \left(x+y\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow \cos xy\frac{d}{dx}\left(xy\right)-\sin \left(x+y\right)\frac{d}{dx}\left(x+y\right)=0 \\ \Rightarrow \cos xy\left[x\frac{dy}{dx}+y\frac{d}{dx}\left(x\right)\right]-\sin \left(x+y\right)\left[1+\frac{\mathit{d}y}{\mathit{d}x}\right]=0 \\ \Rightarrow \cos xy\left[x\frac{\mathit{d}y}{\mathit{d}x}+y\left(1\right)\right]-\sin \left(x+y\right)-\sin \left(x+y\right)\frac{\mathit{d}y}{\mathit{d}x}=0 \\ \Rightarrow x\cos xy\frac{\mathit{d}y}{\mathit{d}x}+y\cos xy-\sin \left(x+y\right)-\sin \left(x+y\right)\frac{\mathit{d}y}{\mathit{d}x}=0 \\ \Rightarrow \left[x\cos xy-\sin \left(x+y\right)\right]\frac{\mathit{d}y}{\mathit{d}x}=\left[\sin \left(x+y\right)-y\cos xy\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\left[\frac{\sin \left(x+y\right)-y\cos xy}{x\cos xy-\sin \left(x+y\right)}\right] \end{gathered}$$